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Cisco EIGRP 路由度量值计算实验

2013-12-20 17:19 浏览:

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1、如图搭建拓扑,做基本配置,确保直连网络连通性。

2、配置EIGRP

R1(config)#router eigrp 100

R1(config)#no au

R1(config)#network x.x.x.x x.x.x.x

3、配置完成EIGRP,测试全网的连通性。

4、EIGRP的度量值计算公式为:

256*{K1(10^7/带宽)+K2(10^7/带宽)/(256-负载)+K3(延迟)+K5/(可靠性+K4)}
默认情况下,K1和K3是1,其他的K值都是0.

所以通常情况下,度量值=256×(10^7/最小带宽+累积延时)

5、接口带宽和延迟通过“show interface”查看

 

R4#show int f 2/0

FastEthernet2/0 is up, line protocol is up

Hardware is AmdFE, address is cc03.1494.0020 (bia cc03.1494.0020)

Internet address is 24.24.24.4/24

MTU 1500 bytes, BW 100000 Kbit, DLY 100 usec,

reliability 255/255, txload 1/255, rxload 1/255

Encapsulation ARPA, loopback not set

Keepalive set (10 sec)

Full-duplex, 100Mb/s, 100BaseTX/FX

----------------------------------------------------------------------------------

R1#show int loopback 0

Loopback0 is up, line protocol is up

Hardware is Loopback

Internet address is 1.1.1.1/24

MTU 1514 bytes, BW 8000000 Kbit, DLY 5000 usec,

reliability 255/255, txload 1/255, rxload 1/255

Encapsulation LOOPBACK, loopback not set

---------------------------------------------------------------------------------

R2#show int f 0/0

FastEthernet0/0 is up, line protocol is up

Hardware is AmdFE, address is cc01.1494.0000 (bia cc01.1494.0000)

Internet address is 12.12.12.2/24

MTU 1500 bytes, BW 100000 Kbit, DLY 100 usec,

reliability 255/255, txload 1/255, rxload 1/255

Encapsulation ARPA, loopback not set

---------------------------------------------------------------------------------

R4#show ip rout

Codes: C - connected, S - static, R - RIP, M - mobile, B - BGP

D - EIGRP, EX - EIGRP external, O - OSPF, IA - OSPF inter area

N1 - OSPF NSSA external type 1, N2 - OSPF NSSA external type 2

E1 - OSPF external type 1, E2 - OSPF external type 2

i - IS-IS, su - IS-IS summary, L1 - IS-IS level-1, L2 - IS-IS level-2

ia - IS-IS inter area, * - candidate default, U - per-user static route

o - ODR, P - periodic downloaded static route

Gateway of last resort is not set

34.0.0.0/24 is subnetted, 1 subnets

C 34.34.34.0 is directly connected, FastEthernet3/0

/ 1.///0.0.0/24 is subnetted, 1 subnets

D 1.1.1.0 [90/158720] via 24.24.24.2, 01:09:05, FastEthernet2/0

2.0.0.0/24 is subnetted, 1 subnets

D 2.2.2.0 [90/156160] via 24.24.24.2, 00:26:13, FastEthernet2/0

3.0.0.0/24 is subnetted, 1 subnets

D 3.3.3.0 [90/156160] via 34.34.34.3, 01:08:54, FastEthernet3/0

4.0.0.0/24 is subnetted, 1 subnets

C 4.4.4.0 is directly connected, Loopback0

23.0.0.0/24 is subnetted, 1 subnets

D 23.23.23.0 [90/30720] via 34.34.34.3, 01:09:06, FastEthernet3/0

[90/30720] via 24.24.24.2, 01:09:06, FastEthernet2/0

24.0.0.0/24 is subnetted, 1 subnets

C 24.24.24.0 is directly connected, FastEthernet2/0

12.0.0.0/24 is subnetted, 1 subnets

D 12.12.12.0 [90/30720] via 24.24.24.2, 01:09:07, FastEthernet2/0

---------------------------------------------------------------------------------

根据以上信息,计算R4到达1.1.1.0网络的度量值

[10^7/100000+(100+100+5000)/10]*256=158720

 

100000是到达1.1.1.0网络的链路的最小带宽,lookback口为8000000;

100+100+5000是R4 F2/0,R2 F0/0, r1 lookback 0口的延迟之和

 

计算R4到达2.2.2.0网络的度量值

[10^7/100000+(100+5000)/10]*256=156160

 

实验结束